Practise › Questions › Volumes of revolution
Volumes of revolution questions
Spin a curve round an axis and it sweeps out a solid. Slice the solid into discs and integration adds them up.
7 original questions · 26 marks · the volumes of revolution notes · Further calculus
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Write down the formula for the volume generated when the region under y = f(x) between x = a and x = b is rotated fully about the x-axis, and state why the y is squared.
Worked answer
V = π∫y² dx from a to b. Each strip sweeps a disc of radius y and thickness dx, and a disc's volume is π times radius squared times thickness. B1 for the formula, B1 for the disc explanation.The region under y = 3 from x = 0 to x = 4 is rotated about the x-axis. Find the volume by integration, and check it against a known solid.
Worked answer
V = π∫9 dx from 0 to 4 = 36π. The solid is a cylinder of radius 3 and length 4, and πr²h = π × 9 × 4 = 36π agrees. M1 for π∫9 dx, A1 for 36π, B1 for the cylinder check. Constant y is the quickest sanity check the formula ever gets.The region under y = x² from x = 0 to x = 2 is rotated fully about the x-axis. Find the exact volume.
Worked answer
V = π∫x⁴ dx from 0 to 2 = π[x⁵/5] = 32π/5. Square first, integrate second. M1 for squaring y, A1 for the integrand, M1 for integrating, A1 for 32π/5. The integrand is x⁴, not x², and the power-rule slip of integrating y before squaring is the topic's classic error.The region bounded by y = x³, the y-axis and y = 8 is rotated about the y-axis. Find the exact volume.
Worked answer
Radius is x with x³ = y, so x² = y2/3 and V = π∫y2/3 dy from 0 to 8 = π[3y5/3/5] = π × 3 × 32/5 = 96π/5. M1 for x2 in terms of y, A1 for the integrand, M1 for integrating with the y-limits, A1 for 96π/5. About the y-axis everything transposes. The limits become y-limits, and x² has to be expressed in terms of y.The region between y = x and y = x² from x = 0 to x = 1 is rotated about the x-axis. Find the volume of the solid formed.
Worked answer
Outer radius x, inner radius x²: V = π∫(x² − x⁴) dx from 0 to 1 = π(1/3 − 1/5) = 2π/15. M1 for the difference of squares, A1 for the integrand, M1 for integrating, A1 for 2π/15. Subtract the squares, never square the difference: π∫(x − x²)² dx describes a different and wrong solid.Explain why rotating the region between two curves uses π∫(outer² − inner²) dx rather than π∫(outer − inner)² dx.
Worked answer
Each cross-section is an annulus, a disc of the outer radius with a disc of the inner radius removed, of area π(outer² − inner²). Squaring the difference instead computes a disc whose radius is the gap between the curves, which is a different region rotated about a different axis. B1 for the annulus, B1 for its area, B1 for what squaring the difference computes instead.A curve has parametric equations x = t², y = 2t for 0 ≤ t ≤ 2. The region bounded by the curve, the x-axis and the line x = 4 is rotated through 2π about the x-axis. Find the exact volume of the solid formed.
Worked answer
The formula V = π∫y² dx still applies, with dx rewritten in terms of t. Here dx/dt = 2t, so dx = 2t dt, and y² = 4t². The limits transform as well, since x = 0 gives t = 0 and x = 4 gives t = 2. So V = π∫4t² × 2t dt from 0 to 2 = π∫8t³ dt = π[2t⁴] from 0 to 2 = 32π. M1 for dx = 2t dt, B1 for the new limits, M1 for y² = 4t², A1 for π∫8t³ dt, M1 for integrating, A1 for 32π. As a check, eliminating t gives y² = 4x, and π∫4x dx from 0 to 4 = π[2x²] = 32π as well. Leaving the limits in x while integrating with respect to t is the error that costs most marks on this type.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise volumes of revolution one question at a time
The player marks nothing for you. It shows one question, waits, then shows the worked answer so you can mark yourself, and brings a question back sooner when it went badly.